I'm having trouble understanding Laravel 5's elixir pathing. In my project, I have multiple css files (bootstrap, plugins, theme etc) and javascript files stored under:
resources/assets/css/<my css files>
resources/assets/js/<my javascript files>
I simply want to combine and version all styles and scripts and place them in my public directory. I believe the default output directly is:
public/build/css/app-xxxxxxxx.css
public/build/js/app-xxxxxxxx.js
(Where xxxxxxxx is the checksum using the version method)
What should my gulpfile.js look like to achieve this? Thanks!
You can use full path on the name or set the third parameter as default path. Examples (works with scripts or css):
mix.stylesIn('resources/assets/css', 'public/css/all.css');
Since there is a bug where you can't use the output of a somethingAll to concatenate with something else, I use this instead (note the wildcard):
mix.scripts(['maskedinput.js',
'blockui.js',
'user/*.js'],
'public/js/user.js', 'resources/assets/js/');
First parameter is the input files, second is the output file, third is the input's default path.
To version just call it on the file path.
mix.version("public/js/user.js");
Related
I'm trying to use Atom to run a Lua script. However, when I try to load files via the require() command, it always says it's unable to locate them. The files are all in the same folder. For example, to load utils.lua I have tried
require 'utils'
require 'utils.lua'
require 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua'
require 'D:\\Users\\Mike\\Dropbox\\Lua Modeling\\utils.lua'
require 'D:/Users/Mike/Dropbox/Lua Modeling/utils.lua'
I get errors like
Lua: D:\Users\Mike\Dropbox\Lua Modeling\main.lua:12: module 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' not found:
no field package.preload['D:\Users\Mike\Dropbox\Lua Modeling\utils.lua']
no file '.\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua\init.lua'
no file 'D:\Program Files (x86)\Lua\5.1\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
The messages says on the first line that 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' was not found, even though that is the full path of the file. What am I doing wrong?
Thanks.
The short answer
You should be able to load utils.lua by using the following code:
require("utils")
And by starting your program from the directory that utils.lua is in:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
The long answer
To understand what is going wrong here, it is helpful to know a little bit about how require works. The first thing that require does is to search for the module in the module path. From Programming in Lua chapter 8.1:
The path used by require is a little different from typical paths. Most programs use paths as a list of directories wherein to search for a given file. However, ANSI C (the abstract platform where Lua runs) does not have the concept of directories. Therefore, the path used by require is a list of patterns, each of them specifying an alternative way to transform a virtual file name (the argument to require) into a real file name. More specifically, each component in the path is a file name containing optional interrogation marks. For each component, require replaces each ? by the virtual file name and checks whether there is a file with that name; if not, it goes to the next component. The components in a path are separated by semicolons (a character seldom used for file names in most operating systems). For instance, if the path is
?;?.lua;c:\windows\?;/usr/local/lua/?/?.lua
then the call require"lili" will try to open the following files:
lili
lili.lua
c:\windows\lili
/usr/local/lua/lili/lili.lua
Judging from your error message, your Lua path seems to be the following:
.\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?\init.lua;D:\Program Files (x86)\Lua\5.1\?.lua
To make that easier to read, here are each the patterns separated by line breaks:
.\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?\init.lua
D:\Program Files (x86)\Lua\5.1\?.lua
From this list you can see that when calling require
Lua fills in the .lua extension for you
Lua fills in the rest of the file path for you
In other words, you should just specify the module name, like this:
require("utils")
Now, Lua also needs to know where the utils.lua file is. The easiest way is to run your program from the D:\Users\Mike\Dropbox\Lua Modeling folder. This means that when you run require("utils"), Lua will expand the first pattern .\?.lua into .\utils.lua, and when it checks that path it will find the utils.lua file in the current directory.
In other words, running your program like this should work:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
An alternative
If you can't (or don't want to) change your working directory to run the program, you can use the LUA_PATH environment variable to add new patterns to the path that require uses to search for modules.
set LUA_PATH=D:\Users\Mike\Dropbox\Lua Modeling\?.lua;%LUA_PATH%;
lua "D:\Users\Mike\Dropbox\Lua Modeling\main.lua"
There is a slight trick to this. If the LUA_PATH environment variable already exists, then this will add your project's folder to the start of it. If LUA_PATH doesn't exist, this will add ;; to the end, which Lua fills in with the default path.
I'm trying to output the compiled CSS to a different directory.
/project/scss/ to /project/css/
I've tried this, but get the error below:
Error: error No such file or directory - /Applications/MAMP/htdocs/project/public/css/style.scss
What are the Arguments and Output paths to refresh exactly for?
I got it working now with the following settings:
The only difference now is the Output paths to refresh field. It's the default now.
$FileParentDir$ contains the path of the parent SCSS file. So in our case it will be /project/scss. In order to make it work like you need, it is necessary to use $ProjectFileDir$ variable or (if you insist to use $FileParentDir$) $FileParentDir$/../ so it would go to an upper level directory.
If you will open any file in Editor and then go to Preferences | Tools | File Watchers > edit your file watcher settings > press Insert Macro button, you will see the list of all the variables with previews for currently opened file so you can see the values and build the arguments accordingly.
Note that Output paths to refresh also should be modified according to the Arguments path.
My purpose is to have an empty hugo application, so, using scripts, I can store list of directories with md files or only md files in an external directory, one level above.
Yes - you can use the contentDir option in your config file, or pass the -c or --contentDir flags to Hugo on the command line.
I'm using Sphinx on Windows.
Most of my documentation is for regular users, but there are some sub-pages with content for administrators only.
So I want to build two versions of my documentation: a complete version, and a second version with the "admin" pages excluded.
I used the exclude_patterns in the build configuration for that.
So far, it works. Every file in every subfolder whose name contains "admin" is ignored when I put this into the conf.py file:
exclude_patterns = ['**/*admin*']
The problem is that I'd like to run the build once to get both versions.
What I'm trying to do right now is running make.bat twice and supply different parameters on each run.
According to the documentation, I can achieve this by setting the BUILDDIR and SPHINXOPTS variables.
So now I have a build.bat that looks like this:
path=%path%;c:\python27\scripts
rem BUILD ADMIN DOCS
set SPHINXOPTS=
set BUILDDIR=c:\build\admin
call make clean
call make html
rem BUILD USER DOCS
set SPHINXOPTS=-D exclude_patterns=['**/*admin*']
set BUILDDIR=c:\build\user
call make clean
call make html
pause
The build in the two different directories works when I delete the line set BUILDDIR=build from the sphinx-generated make.bat file.
However, the exclude pattern does not work.
The batch file listed above outputs this for the second build (the one with the exclude pattern):
Making output directory...
Running Sphinx v1.1.3
loading translations [de]... done
loading pickled environment... not yet created
Exception occurred:
File "C:\Python27\lib\site-packages\sphinx-1.1.3-py2.7.egg\sphinx\environment.
py", line 495, in find_files
['**/' + d for d in config.exclude_dirnames] +
TypeError: coercing to Unicode: need string or buffer, list found
The full traceback has been saved in c:\users\myusername\appdata\local\temp\sphinx-err-kmihxk.log, if you want to report the issue to the developers.
Please also report this if it was a user error, so that a better error message can be provided next time.
Either send bugs to the mailing list at <http://groups.google.com/group/sphinx-dev/>,
or report them in the tracker at <http://bitbucket.org/birkenfeld/sphinx/issues/>.
What am I doing wrong?
Is the syntax for exclude_patterns in the sphinx-build command line different than in the conf.py file?
Or is there a better way to build two different versions in one step?
My first thought was that this was a quoting issue, quoting being notoriously difficult to get right on the Windows command line. However, I wasn't able to come up with any combination of quoting that changed the behavior at all. (The problem is easy to replicate)
Of course it could still just be some quoting issue I'm not smart enough to figure out, but I suspect this is a Sphinx bug of some kind, and hope you will report it to the Sphinx developers.
In the meantime, here's an alternate solution:
quoting from here:
There is a special object named tags available in the config file. It can be used to query and change the tags (see Including content based on tags). Use tags.has('tag') to query, tags.add('tag') and tags.remove('tag') to change
This allows you to essentially pass flags into the conf.py file from the command line, and since the conf.py file is just Python, you can use if statements to set the value of exclude_patterns conditionally based on the tags you pass in.
For example, you could pass Sphinx options like:
set SPHINXOPTS=-t foradmins
to pass the "foradmins" tag, and then check for it in your conf.py like so:
exclude_patterns = blah
if tags.has('foradmins'):
exclude_patterns = []
That should allow you to do what you want. Good Luck!
I have a conf file that is of the format:
name=value
What I want to do is using a template, generate a result based on some values in another file.
So for example, say I have a file called PATHS that contains
CONF_DIR=/etc
BIN_DIR=/usr/sbin
LOG_DIR=/var/log
CACHE_DIR=/home/cache
This PATHS file gets included into a Makefile so that when I call make install the paths are created and built applications and conf files copied appropriately.
Now I also have a conf file which I want to use as a template.
Say the template contains lines like
LogFile=$(LOG_DIR)/myapp.log
...
Then generate a destination conf that would have
LogFile=/var/log/myapp.log
...
etc
I think this can be done with a sed script, but I'm not very familiar with sed and regular expression syntax. I will accept a shell script version too.
You should definitely go with autoconf here, whose very job is to do this. You'll have to write a conf.in file, wherein all substitutions are marked with #'s, e.g.
prefix=#prefix#
bindir=#bindir#
and write up a configure.ac, which is a shell script that will perform these substitutions for you and create conf. conf is subsequently included in the Makefile. I'd even recommend using a Makefile.in file, i.e. including your snippet in the Makefile.
If you keep to the standard path names, your configure.ac is a four-liner and has the added advantage of being GNU compatible (easy to understand & use).
You may want to consider using m4 as a simple template language instead.